2. If $$ x-i y=\sqrt{\frac{a-i b}{c-i d}} $$ prove that $$ \left(x^{2}+y^{2}\right)=\frac{a^{2}+b^{2}}{c^{2}+d^{2}} $$

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To prove that $$ x^2 + y^2 = \frac{a^2 + b^2}{c^2 + d^2} $$ given that $$ x - i y = \sqrt{\frac{a - i b}{c - i d}} $$, follow these steps:

1. **Understand the Given Equation:**
   $$
   x - i y = \sqrt{\frac{a - i b}{c - i d}}
   $$
   Here, $$ x $$ and $$ y $$ are real numbers, and $$ i $$ is the imaginary unit.

2. **Compute the Modulus on Both Sides:**
   The modulus (or absolute value) of a complex number $$ z = p + i q $$ is $$ |z| = \sqrt{p^2 + q^2} $$.

Take the modulus of both sides:
   $$
   |x - i y| = \left| \sqrt{\frac{a - i b}{c - i d}} \right|
   $$

3. **Simplify the Right Side:**
   The modulus of a square root is the square root of the modulus:
   $$
   |x - i y| = \sqrt{ \left| \frac{a - i b}{c - i d} \right| }
   $$

4. **Compute the Modulus of the Fraction:**
   $$
   \left| \frac{a - i b}{c - i d} \right| = \frac{|a - i b|}{|c - i d|}
   $$
   Calculate each modulus:
   $$
   |a - i b| = \sqrt{a^2 + b^2}
   $$
   $$
   |c - i d| = \sqrt{c^2 + d^2}
   $$
   Therefore:
   $$
   \left| \frac{a - i b}{c - i d} \right| = \frac{\sqrt{a^2 + b^2}}{\sqrt{c^2 + d^2}}
   $$

5. **Combine the Results:**
   $$
   |x - i y| = \sqrt{ \frac{\sqrt{a^2 + b^2}}{\sqrt{c^2 + d^2}} } = \sqrt{ \frac{a^2 + b^2}{c^2 + d^2} }
   $$

6. **Square Both Sides to Eliminate the Square Root:**
   $$
   |x - i y|^2 = \frac{a^2 + b^2}{c^2 + d^2}
   $$
   Since $$ |x - i y|^2 = x^2 + y^2 $$, we have:
   $$
   x^2 + y^2 = \frac{a^2 + b^2}{c^2 + d^2}
   $$

**Conclusion:**
$$
x^2 + y^2 = \frac{a^2 + b^2}{c^2 + d^2}
$$
$$ x^2 + y^2 = \frac{a^2 + b^2}{c^2 + d^2} $$

If prove that

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